48035-8.1-10E
AID:114681 | 06/05/2018
We have an initial population which is 200. Based on this initial population, after 5 hours,
the new population is 200*3 600. Now, we have to find out, at which hour or after
how many hours, will the population of the bacterial culture become 20000. Also we
need to derive an equation to find the population at any point of time.
If we let y(t) represent the number of bacteria in a culture at time t, then the rate of
change of population with respect to time is
y(t) , we have
y' (t) . Thus, since y' (t) is proportional to
y' (t) k y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
Considering the equation
We have
y(t) Aekt ------ (i)
y
(0)
200
Based on (i), we have y(0) 200 = Aek*0 = A (since e0 1 )
Therefore, 200= A
Subsequent y(5) 600 200ek*5 , based on the problem; replacing
t=5,y(t)=600, A=200
Now we get
3
e
k*5
,solving the above equation
Applying natural logs on both sides, ln 3 5k
Thus k ln 3
5
From the problem, substituting the values in equation (i), we get,
ln 3t
y(t) 200e 5 ------- (ii)
ln 3t
ln 2
20000 200e 5
, substituting
k
,
y
(
t
)
6000
4
So, we get
Applying logs on both sides,
Finally ,
ln 3t
100 e 5
ln100 ln 3 t
5
t
(ln100)*5
4.605*5
20.958
hrs
ln 3 1.0986
Thus, ln 3t
y(t) 200e 5 is the required equation for the population at any time.
Also, at 20.96th hr, the population will reach 20000.
48035-8.1-11E
AID:114681 | 06/05/2018
We have an initial population which is 400. Based on this initial population, after 1 hour,
the new population is 800. Now, we have to find out, what would be the population after
10hours. Also, we need to derive an equation to find the population at any point of time.
If we let y(t) represent the number of bacteria in a culture at time t, then the rate of
change of population with respect to time is
y(t) , we have
y' (t) . Thus, since y' (t) is proportional to
y' (t) k y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
Considering the equation
We have
y(t) Aekt ------(i) 2.718
y
(0)
400
Based on (i), we have y(0) 400 = Aek*0 = A (since e0 1 )
Therefore, 400= A
Subsequent , y(1) 800 400ek*1 400ek based on the problem; replacing
Now we get 2 ek
t=1,y(t)=800, A=400
,solving the above equation
Applying natural logs on both sides, ln 2 k
Thus k ln 2
From the problem, substituting the values in equation (i), we get,
y(t) 400e(ln2)*t ------(ii)
So, we get,y(t)=y(10)
y(t) 400e(ln2)*10 , substituting k ln 2 ,t=10
y(t) 400e0.693*10 400e6.931 409406.794
Thus,
y(t) 400e(ln2)*t is the required equation for the population at any time.
Also, at 10th hr, the population will reach 409406.794
48035-8.1-12E
AID:114681 | 06/05/2018
We have an initial population which is100. Based on this initial population, after 2 hour,
the new population is 400 . Now, we have to find out, what would be the population after
8 hours. Also, we need to derive an equation to find the population at any point of time.
If we let y(t) represent the number of bacteria in a culture at time t, then the rate of
change of population with respect to time is
y(t) , we have
y' (t) . Thus, since y' (t) is proportional to
y' (t) k y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
Considering the equation
We have
y(t) Aekt ------ (i)
y
(0)
100
Based on (i), we have y(0) 100 = Aek*0 = A (since e0 1 )
Therefore, 100 A
Subsequent y(2) 400 100ek*2 , based on the problem; replacing
t 2, y t 400, A 100
Now we get 4 e2k ,solving the above equation
Applying natural logs on both sides, 1 ln 4 k
2
Thus k 1 ln 4
2
From the problem, substituting the values in equation (i), we get,
y(t) 100e
(ln 4)*t*1
2
(ln 4)*8*1
------(ii)
1
So, we get, y t
y 8
y(t) 100e 2 , substituting k ln 4 , t 8
2
y(t) 100e5.545 100* 255.954578243 25595.4578243
Thus, y(t) 100e (ln 4)*t*1
2 is the required equation for the population at any time.
Also, at 8th hr, the population will reach 25595.4
48035-8.1-13E
AID:114681 | 06/05/2018
We have an initial information where in it says that a bacterial culture grows
exponentially with a growth constant 0.44hour1 . We need to find the doubling time,
which is the time it takes for the bacterial culture to grow to its double size.
If we let y(t) represent the number of bacteria in a culture at time t, then the rate of
change of population with respect to time is
y(t) , we have
y' (t) . Thus, since y' (t) is proportional to
y' (t) k y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
Considering the equation
We have
y(t) Aekt ------ (i)
y(t) Ae0.44t
Let the initial population be 100so
growth constant is k 0.44
y(0) 100 and again according to the problem,
Based on (i), we have y(0) 100 = Ae0.44*0 = A (since e0 1 )
Therefore, 100 A ----- (ii)
Based on the problem, we need to find out the time when the bacterial population
becomes double that is 200
Subsequent y(t) 200 100e0.44*t , based on the problem; replacing
y t 200, A 100 from(ii)
Now we get
2
e
0.44t
,solving the above equation
Applying natural logs on both sides, ln 2 0.44t
Thus t ln 2
0.44
1.575hrs
Thus, t ln 2
0.44
1.575hrs which is the required equation for the population to grow
double it's size. Hence the doubling time is 1.575hrs .
48035-8.1-14E
AID:114681 | 06/05/2018
We have an initial information where in it says that a bacterial culture grows
exponentially with a growth constant 0.12hour1 . We need to find the doubling time,
which is the time it takes for the bacterial culture to grow to its double size.
If we let y(t) represent the number of bacteria in a culture at time t, then the rate of
change of population with respect to time is
y(t) , we have
y' (t) . Thus, since y' (t) is proportional to
y' (t) k y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
Considering the equation
We have
y(t) Aekt ------ (i)
y(t) Ae0.12t
Let the initial population be 100so
growth constant is k 0.44
y(0) 100 and again according to the problem,
Based on (i), we have y(0) 100 = Ae0.12*0 = A (since e0 1 )
Therefore, 100 A ----- (ii)
Based on the problem, we need to find out the time when the bacterial population
becomes double that is 200
Subsequent y(t) 200 100e0.12*t , based on the problem; replacing
y t 200, A 100 from(ii)
Now we get
2
e
0.12t
,solving the above equation
Applying natural logs on both sides, ln 2 0.12t
Thus t ln 2
0.12
5.7762hrs
Thus, t ln 2
0.12
5.7762hrs which is the required equation for the population to grow
double it's size. Hence the doubling time is 5.7762hrs
48035-8.1-15E
AID:114681 | 06/05/2018
We have an initial information wherein a population of E.coli doubles every 20mins. The
initial population y(0) , according to the problem, is 108 .Therefore, y(0) 108 .Thus the
population after 20mins become y(20) 2 y(0) 2*108 .
Here’s a situation, the population of E.coli, sized 108 , starts growing till time, T minutes.
At the T th minute, a treatment of the infection was applied which reduced the size by
90%, and got the population of E.coli back of size 108 .
We need to find the value of T
If we let y(t) represent the size of E.coli in a culture at time t, then the rate of change of
population with respect to time is
y' (t) k
y' (t) . Thus, since
y(t )
y' (t) is proportional to y(t) , we have
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
Considering the equation
We have
Thus
y(t) Aekt ------ (i)
y(0) 108 Aek*0 A
A 108 ------
(ii)
After 20mins, the size doubles y(20) 2 y(0) 2*108 .
y(t) y(20) 2 y(0) 2*108 = Aek*20 from (i)
From (ii), we get 2*108 108 e20k ------ (iv)
108
2* 8
10 e20k
Therefore, applying natural logs on both the sides
ln 2 k ------
(v)
20
The population of E.coli, sized 108 , starts growing till time, T minutes. At the T th minute,
a treatment of the infection was applied which reduced the size by 90%, and got the
population of E.coli back of size 108 .
This means that finally 10% of size is 108 .
Therefore,
So,
10% 108
10
8
100% 10
*100 ------ (iii)
108 *100 ln 2T
Thus, the size at time, T is y(T )
10
8
e
20
10 , we get this from (iii),(iv)&(v)
Therefore, 108 *100
10*108
ln 2T
e 20
Applying natural logs on both sides, we get,
ln10 ln 2 T
20
T 20 ln10
ln 2
66.4min s
Thus, T 20 ln10
ln 2 66.4min s which is the required value for the problem.
48035-8.1-16E
AID:114681 | 06/05/2018
We have an initial information wherein in the year 1950, it took 109 acres of arable land
to sustain world population. Hence, we can consider y(0) 109 ------ (i)
Now, in the year 1980, that is 30 years later, it takes 2 *109 acres of arable land to support
world population. Hence, we have to consider
equation (i)
y(30) 2*109 comparing this condition to
Finally, the total arable land available is 3.2 *109 .
Thus, let's consider y(t) 3.2*109 ------
(vi)
We need to find the time, t , which would be required to fill up the complete 3.2 *109
acres of arable land available to support world population.
If we let y(t) represent the number of acres at time t, then the rate of change of acres of
land with respect to time is y' (t) . Thus, since y' (t) is proportional to y(t) , we have
y' (t) k y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
Considering the equation
We have
Thus
y(t) Aekt ------ (ii)
y(0) 109 Aek*0 A
A 109 ------
(iii)
After 30 years, the size doubles y(30) 2 y(0) 2*109 .
From (iii), we get
y(30) 2 y(0) 2*109 Aek*30
2*109 109 e30k
109
2* 9
10 e30k
Therefore, applying natural logs on both the sides
ln 2 k ------
(iv)
30
From (vi), we get, ln 2 t
y(t) 3.2*109 =109 e 30
,substituting from(iv)&(iii)
3.2*
10
9
10
9
ln 2t
e 30
Applying natural logs on both the sides,
Hence,
ln 3.2 ln 2 t
30
t
30 ln 3.2
30 *1.16315
50.34
50
ln 2 .69315
Thus, t 50 years that is, in the year 1950 50 2000 the complete 3.2 *109 acres of
arable land available to support world population would be occupied.
48035-8.1-17E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that some quantity is increasing
exponentially with growth rate r. We need to prove that doubling time is ln 2 .
r
If we let y(t) represent the ‘some quantity’ at time t, then the rate of change of ‘some
quantity’ with respect to time is
y' (t) r
y' (t) . Thus, since
y(t )
y' (t) is proportional to y(t) , we have
For some constant of proportionality r (the growth constant)
Now, from the above equation, we can derive,
y(t) Aert , where A=constant, r =growth constant t=time
Considering the equation
We have
y(t) Aert ------ (ii)
y(0) Aer*0 A ------ (i)
The quantity that represents after doubling time is
y(t) 2 y(0) 2A Aer*t , we get the values from (i)and (ii)
Therefore 2 A ert
A
Thus, applying natural logs on both sides, we get
ln 2 rt
Finally t ln 2
r
Thus we have a final proof that, doubling time is t ln 2
r
48035-8.1-18E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that some quantity is increasing
exponentially with growth rate r. We need to prove that half time is ln 2 .
r
If we let y(t) represent the ‘some quantity’ at time t, then the rate of change of ‘some
quantity’ with respect to time is
y' (t) r
y' (t) . Thus, since
y(t )
y' (t) is proportional to y(t) , we have
For some constant of proportionality r (the growth constant)
Now, from the above equation, we can derive,
y(t) Aert , where A=constant, r =growth constant t=time
Considering the equation
We have
y(t) Aert ------ (ii)
y(0) Aer*0 A ------ (i)
The quantity that represents after half time is
y(t) 1 y(0) 1 A Aer*t , we get the values from (i)and
2 2
(ii)
Therefore 1 * A ert
2
A
Thus, applying natural logs on both sides, we get
ln 1 rt , now we know that ln 1 ln 2
Finally
2 2
t ln 2
r
Thus we have a final proof that, doubling time is t ln 2
r
Half life in this problem is 1 times the doubling time in exercise 17
48035-8.1-19E
AID:114681 | 06/05/2018
We have an initial information that the radioactive element iodine-131 has a decay
constant of 1.3863day1 . Now, we have to find out, it’s half life which is equal to the
half of initial condition= 1 y(0) , considering
2
y(0) as the initial condition.
If we let y(t) represent the number of iodine molecules in a culture at time t, then the rate
of change of iodine molecules with respect to time is y' (t) . Thus, since y' (t) is
proportional to y(t) , we have
y' (t) k
y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
Considering the equation y(t) Aekt ------ (i)
Considering y(0) as the initial condition.
Based on (i), we have
Therefore
y(0) Aek*0 = A (since e0 1 )
y(t) 1 y(0) A Ae1.3863*t
2 2
Subsequently, crossing A from both sides, we get
1
e
1.3863*t
2
Now we get, after applying logs on both the sides and also considering ln 1 ln 2 from
2
the logarithmic formula
ln 1
ln 2
1.3863t
2
Thus
t
ln 2
1.3863
Thus, t 0.5day , which is the half-life of radioactive element, iodine.
0.5
day
48035-8.1-1E
AID:114681 | 06/05/2018
y
We have to solve the equation y 4 y with y(0) 2 and check our solution.
We can consider that y(x) 2 . In this regard, we can consider x=0 and y=2
So our equation is:
This also means
Now, we can get
Thus, we can get
y 4 y
dy 4 y
dx
dy 4 ydx
dy 4dx
y
Now integrating dy 4dx , both the sides
ln y c1 4x c2 , c1 & c2 are the constants
ln y 4x (c2 c1)
ln y 4x c , replacing the constant with just c
e
ln y
e4xc , introducing exponentials on both the sides
y e
4x
e
c
, changes happen based on the basic formulas
y Ae4x , considering the constant as A --------------- (i)
y Ae4x , this is our final equation
2 Ae4*0 , replacing the values from our problem i.e. x=0 y=2
2 A
Finally considering the value of our constant 2 A
From the equation (i), we get, y 2e4x
Thus, y 2e4x is a solution of the differential equation that satisfies the initial indicated
condition.
48035-8.1-20E
AID:114681 | 06/05/2018
We have an initial information that the radioactive element iodine-137 has a decay
constant of 0.023year1 . Now, we have to find out, it’s half life which is equal to the
half of initial condition= 1 y(0) , considering
2
y(0) as the initial condition.
If we let y(t) represent the number of iodine molecules in a culture at time t, then the rate
of change of iodine molecules with respect to time is y' (t) . Thus, since y' (t) is
proportional to y(t) , we have
y' (t) k
y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
Considering the equation y(t) Aekt ------ (i)
Considering y(0) as the initial condition.
Based on (i), we have
Therefore
y(0) Aek*0 = A (since e0 1 )
y(t) 1 y(0) A Ae0.023*t
2 2
Subsequently, crossing A from both sides, we get
1
e
0.023*t
2
Now we get, after applying logs on both the sides and also considering ln 1 ln 2 from
2
the logarithmic formula
Thus
ln 1 ln 2 0.023t
2
t ln 2 30.14 / year
0.023
Thus, t 30.14 / year , which is the half-life of radioactive element, iodine.
48035-8.1-2E
AID:114681 | 06/05/2018
We have to solve the equation with y 3 y with y(0) 2 and check our solution.
We can consider that y(x) 2 . In this regard, we can consider
x
0 &
y
2
So our equation is:
This also means
Now, we can get
Thus, we can get
y 3 y
dy 3y
dx
dy
3
ydx
dy 3dx
y
Now integrating dy 3dx , both the sides
dx
ln y c1 3 x c2 , c1 & c2 are the constants
ln y =3 x +( c2 c1 )
ln y 3x c , replacing the constant with just c
eln y = e3xc , introducing exponentials on both the sides
y = e3x ec , changes happen based on the basic formulas
y = Ae3x , considering the constant as A --------------- (i)
y = A e3x , this is our final equation
2 Ae4*0 , replacing the values from our problem i.e. x=0 y= 2
2=A
Finally considering the value of our constant A= 2
From the equation (i), we get, y = 2 e3x
Thus, y = 2 e3x is a solution of the differential equation that satisfies the initial indicated
condition.
48035-8.1-3E
AID:114681 | 06/05/2018
y
We have to solve the equation y 3y with y(0)=5 and check our solution.
We can consider that y(x) 5 . In this regard, we can consider
x
0,
y
5
So our equation is:
This also means
Now, we can get
Thus, we can get
y
3
y
dy
3
y
dx
dy
3
ydx
dy 3dx
y
Now integrating dy 3dx , both the sides
ln y c1 3x c2 , c1 & c2 are the constants
ln y c1 3x (c2 c1)
ln y 3x c , replacing the constant with just c
e
ln y
e3xc , introducing exponentials on both the sides
y e3xec , changes happen based on the basic formulas
y Ae3x , considering the constant as A --------------- (i)
y Ae3x , this is our final equation
5 Ae3*0 , replacing the values from our problem i.e. x=0 y=5
5=A
Finally considering the value of our constant A=5
From the equation (i), we get,
y 5e
3x
Thus, y 5e3x is a solution of the differential equation that satisfies the initial indicated
condition.
48035-8.1-4E
AID:114681 | 06/05/2018
y
We have to solve the equation y 2 y with
y 0
6 and check our solution.
We can consider that y(x) 6 . In this regard, we can consider x 0, y 6
So our equation is:
This also means
Now, we can get
Thus, we can get
y 2 y
dy 2 y
dx
dy 2 ydx
dy 2dx
y
Now integrating dy 2dx , both the sides
ln y c1 2x c2 , c1 & c2 are the constants
ln y c1 2x (c2 c1)
ln y 2x c , replacing the constant with just c
e
ln y
e2xc , introducing exponentials on both the sides
y e2xec , changes happen based on the basic formulas
y Ae2x , considering the constant as A --------------- (i)
y Ae
2x
, this is our final equation
6 Ae2*0 , replacing the values from our problem i.e. x=0 y=5
6 A
Finally considering the value of our constant A 6
From the equation (i), we get, y 6e2x
Thus, y 6e2x is a solution of the differential equation that satisfies the initial indicated
condition.
48035-8.1-5E
AID:114681 | 06/05/2018
We have to solve the equation y 2 y with
y 1
2 and check our solution.
We can consider that y(x) 2 . In this regard, we can consider x 1, y 2
So our equation is:
This also means
Now, we can get
Thus, we can get
y 2 y
dy 2 y
dx
dy 2 ydx
dy 2dx
y
Now integrating , both the sides
ln y c1 2x c2 , c1 & c2 are the constants
ln y c1 2x (c2 c1)
ln y 2x c , replacing the constant with just c
e
ln y
e2xc , introducing exponentials on both the sides
y e
2x
e
c
, changes happen based on the basic formulas
y Ae2x , considering the constant as A --------------- (i)
y Ae2x , this is our final equation
2 Ae2*1 , replacing the values from our problem i.e. x=1 y=2
2
Ae
2
A
2
e
2
Finally considering the value of our constant,
A
2
e
2
From the equation (i), we get,
y
2
e
2 x
2
e
2 x
2
2
e
2( x
1)
e
2
Thus, y 2e2(x1) is a solution of the differential equation that satisfies the initial indicated
condition.
48035-8.1-6E
AID:114681 | 06/05/2018
We have to solve the equation y 1y with
y 1
2 and check our solution.
We can consider that y(x) 2 . In this regard, we can consider x 1, y 2
So our equation is:
y
1
y
This also means dy y
dx
Now, we can get dy ydx
Thus, we can get dy dx
y
Now integrating , both the sides
ln y c1 1x c2 , c1 & c2 are the constants
ln y c1 1x (c2 c1)
ln y 1x c , replacing the constant with just c
e
ln y
exc , introducing exponentials on both the sides
y e
1x
e
c
, changes happen based on the basic formulas
y Ae1x , considering the constant as A --------------- (i)
y Ae1x , this is our final equation
2 Ae1*1 , replacing the values from our problem i.e. x=1 y=2
2
Ae
1
A
2
e
1
Finally considering the value of our constant,
A
2
e
1
From the equation (i), we get,
y
2
e
1x
2
e
1x
1
2
e
1( x
1)
e
1
Thus, y 2e
1(x
1)
is a solution of the differential equation that satisfies the initial
indicated condition.
48035-8.1-7E
AID:114681 | 06/05/2018
We have to solve the equation y 3 with
y 0
3and check our solution.
We can consider that y(x) 3 . In this regard, we can consider x 0, y 3
So our equation is:
This also means
Thus, we can get
y
3
dy 3 Now, we can get
dx
dy
3
dx
dy
3
dx
Now integrating
From (i),we get
dy 3dx , both the sides
y c1 3x c2 , c1 & c2 are the constants
y c1 3x (c2 c1)
y 3x c , replacing the constant with just c ----- (i)
y 3* 0 c , considering the value of x as 0-based on the problem
y c 3 , considering the value of y as 3-based on the problem
y 3x 3 , considering c 3
y 3x 3 , this is our final equation
Thus,
y
3
x
3
is a solution of the differential equation that satisfies the initial
indicated condition.
48035-8.1-8E
AID:114681 | 06/05/2018
We have to solve the equation y 2 with
y 0
8 and check our solution.
We can consider that y(x) 8 . In this regard, we can consider x 0, y 8
So our equation is:
This also means
y
2
dy
2
dx
Now, we can get dy 2dx
Thus, we can get
Now integrating
From (i),we get
dy
2
dx
dy 2dx , both the sides
y c1 2x c2 , c1 & c2 are the constants
y 2x (c2 c1)
y 2x c , replacing the constant with just c ----- (i)
y 2*0 c , considering the value of x as 0-based on the problem
y c 8 , considering the value of y as 3-based on the problem
y 2x 8 , considering c 8
y 2x 8 , this is our final equation
Thus, y 2x 8 is a solution of the differential equation that satisfies the initial
indicated condition.
48035-8.1-21E
AID:114681 | 06/05/2018
1
We have an initial information that the half-life of morphine is 3hrs and there’s that
0.4mg of morphine is there in the bloodstream.
Based on the 2nd initial condition, we need to find an equation for the amount of
morphine in the bloodstream at any point of time.
Also, when does the amount of morphine in the bloodstream drop below 0.01mg?
Based on the question, y(t) 0.01 and we need to find the value of t.
If we let y(t) represent the amount of morphine in the bloodstream at any time t, then the
rate of change of amount of morphine with respect to time is y' (t) . Thus, since y' (t) is
proportional to y(t) , we have
y' (t) k
y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
Considering the equation y(t) Aekt ------ (i)
Considering y(0) as the initial condition.
y(0) 0.4 Aek*0 A
A 0.4
Therefore 1 y(0) 0.2 0.4ek*3 (half-life of morphine is 3hrs means
2
that it takes 3hrs to become half of the initial amount(0.4mg) that is 0.2mg)
0.2
e
k*3
0.4
1
e
k*3
2
Applying natural logs on both sides, we get
ln( )
3
k
2
ln 1
Thus, k 2
3
Now, we need to find out the time it takes for the amount of morphine, present originally,
to become 0.01mg
ln 1
2 t
y(t) 0.01 0.4e 3
------(i)
Therefore,
0.01
0.4
ln 1
2 t
e 3
ln 1
2 t
0.025 e 3
Applying the natural logs on both the sides,
ln 1
ln(0.025) 2 t
3
3* ln(0.025) t
ln 0.5
Based on the equation (i), the equation for the amount of morphine at any time is
ln 2 t
0.4e 3 .
Again the time required for the amount of morphine to drop down to 0.01mg is 15.97hrs
t 15.97
48035-8.1-22E
AID:114681 | 06/05/2018
1
We have an initial information that the half-life of morphine is 3hrs and there’s that
0.4mg of morphine is there in the bloodstream.
Based on the 2nd initial condition, we need to find an equation for the amount of
morphine in the bloodstream at any point of time.
Also, when does the amount of morphine in the bloodstream drop below 0.01mg?
Based on the question, y(t) 0.01 and we need to find the value of t.
We have to finally find out the percentage decrease of amount of morphine in the
bloodstream within a day that is in 24 hrs
So here, we need to first get the value of
100 100 *[ y(24)] --------
(iii)
0.4
y(24) , thereby get the final value of
If we let y(t) represent the amount of morphine in the bloodstream at any time t, then the
rate of change of amount of morphine with respect to time is y' (t) . Thus, since y' (t) is
proportional to y(t) , we have
y' (t) k
y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
Considering the equation y(t) Aekt ------ (i)
Considering y(0) as the initial condition.
y(0) 0.4 Aek*0 A
A 0.4
Therefore 1 y(0) 0.2 0.4ek*3 (half-life of morphine is 3hrs means
2
that it takes 3hrs to become half of the initial amount(0.4mg) that is 0.2mg)
0.2
e
k*3
0.4
1
e
k*3
2
Applying natural logs on both sides, we get
ln( )
3
k
2
ln 1
Thus, k 2
3
Now, we need to find out the time it takes for the amount of morphine, present originally,
to become 0.01mg
ln 1
2 t
y(t) 0.01 0.4e 3
------(i)
Therefore,
0.01
0.4
ln 1
2 t
e 3
ln 1
2 t
0.025 e 3
Applying the natural logs on both the sides,
ln 1
ln(0.025) 2 t
3
3* ln(0.025) t
ln 0.5
t
Now, we have to find out the y(24) by replacing t 24 in equation (i)
ln 1
2 *24
y
(24)
0.4
e
3
0.4
e
5.5452
0.00156
Based on equation (iii)
100 100 *[ y(24)] =100 100 *(0.00156) 100 0.3906 99.609
0.4 0.4
Based on the equation (i), the equation for the amount of morphine at any time is
ln 2 t
0.4e 3 .
Again the time required for the amount of morphine to drop down to 0.01mg is 15.97hrs
Thus there’s a percentage decrease of 99.609in the amount of morphine in the blood
within a day
15.97
48035-8.1-23E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that half-life of strontium-90 is 28
years. We need to find out the percentage of strontium is left over in human body, 50
years after being exposed to nuclear explosion.
If we let y(t) represent the ‘some quantity’ at time t, then the rate of change of ‘some
quantity’ with respect to time is y' (t) . Thus, since y' (t) is proportional to y(t) , we have
y' (t) k y(t )
For some constant of proportionality k (the growth constant)
Now, from the above equation, we can derive,
y(t) Aert , where A=constant, k =growth constant t=time
Considering the equation
We have
y(t) Aert ------ (ii)
y(0) Aer*0 A ------ (i)
The quantity that represents after half time is
y(28) 1 y(0) 1 A Aek*28 , we get the values from (i)and
2 2
(ii)
Therefore
1 *
A
e
28*r
2
A
Thus, applying natural logs on both sides, we get
ln 1 28* k , now we know that ln 1 ln 2
Finally
2 2
k ln 2 -------
(iii)
28
Substituting the values of (iii) in (i) and taking up t 50 , we get,
ln 1
2 *50
y
(50)
Ae
28
Ae
1.237763
A
* 0.29003 ------
(iv)
Now, the final result could be got, if we can find
(iv) *100%
(i)
Thus
0.29003A *100% 29%
A
Thus we have a final result as 29%
48035-8.1-24E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that half-life of uranium is
0.7*109 years . We need to find out the quantity that would remain after 100 years,
considering 50gms of uranium are buried at a nuclear site
If we let y(t) represent the ‘some quantity’ at time t, then the rate of change of ‘some
quantity’ with respect to time is y' (t) . Thus, since y' (t) is proportional to y(t) , we have
y' (t) k y(t )
For some constant of proportionality k (the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k =growth constant t=time
Considering the equation
We have
y(t) Aekt ------ (ii)
y(0) Aek*0 A ----- (i)
The quantity that represents after half time is
1
y
(0)
1
A
Ae
k*0.7*10
9 , we get the value
s
from (i)and (ii)
2 2
Therefore
1
*
A
e
k*0.7*10
9
2
A
Thus, applying natural logs on both sides, we get
ln 1 0.7 *109 * k , now we know that ln 1 ln 2
Finally
2
k
ln 2
2
------(iii)
0.7 *109
Substituting the values of (iii) in (i) and taking up t 50 , we get,
ln 1
2 *100
y(100)
Ae
0.7*10
9
Ae
9.9*10
A*0.99999901 ---- (iv)
Now, the final result could be got, if we can find
(iv) *100%
(i)
Thus
0.9999 A *100% 99%
A
Thus we have a final result as 99%
8
48035-8.1-25E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that half-life of carbon 14 is
5730 years . Scientists dating a fossil estimate that 20% of the original amount of
carbon 14 is present. We need to find out the age of the fossil.
If we let y(t) represent the ‘some quantity’ at time t, then the rate of change of ‘some
quantity’ with respect to time is y' (t) . Thus, since y' (t) is proportional to y(t) , we have
y' (t) k y(t )
For some constant of proportionality k (the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k =growth constant t=time
Considering the equation
We have
y(t) Aekt ------ (ii)
y(0) Aek*0 A ----- (i)
The quantity that represents after half time is
1 y(0) 1 A Aek*5730 , we get the values from (i)and (ii)
2 2
Therefore
1 *
A
e
k*5730
2
A
Thus, applying natural logs on both sides, we get
ln 1 5730* k , now we know that ln 1 ln 2
Finally,
2
k ln 2
5730
2
------(iii)
Scientists dating a fossil estimate that 20% of the original amount of carbon 14 is
present.
From (i), we get, A ------(iii) , which is 20% of A
5
Substituting the values of (iii) in (i), we get,
A
ln 1
2 *t
y(t) Ae5730
5 -----(iv)
A * 1
5 A
Applying natural logs on both the sides,
ln 1
ln 1
2 *t
e5730
ln 1 2 *t
5 5730
5730* ln 1
t 5 5730*(1.609) 9222.079 13307.47 years
ln 1
2
0.693 .693
Thus we have a final result as 13307.47 years
48035-8.1-26E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that half-life of carbon 14 is
5730 years . carbon 14 fossil is 1 million years old is present. We need to find out what
percentage of original carbon 14 should remain.
If we let y(t) represent the ‘some quantity’ at time t, then the rate of change of ‘some
quantity’ with respect to time is y' (t) . Thus, since y' (t) is proportional to y(t) , we have
y' (t) k y(t )
For some constant of proportionality k (the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k =growth constant t=time
Considering the equation
We have
y(t) Aekt ------ (ii)
y(0) Aek*0 A ----- (i)
The quantity that represents after half time is
1 y(0) 1 A Aek*5730 , we get the values from (i)and (ii)
2 2
Therefore
1 *
A
e
k*5730
2
A
Thus, applying natural logs on both sides, we get
ln 1 5730* k , now we know that ln 1 ln 2
Finally,
2
k ln 2
5730
2
------(iii)
We need to find out what percentage of original carbon 14 should remain.
ln 1
2 *10
6
From (ii), we get, y(106 ) Ae5730 ------(iv)
We need to find out what percentage of original carbon 14 should remain. In order to
find that out that, we need to get (iv) *100
(i)
Thus
(iv)
(i)
*100 becomes
ln 1
2 *10
6
Ae
5730
A
*100 e
0.00012096809*106
*100 %= 99.9879%
Thus we have a final result as 99.9879%that is what carbon 14 should remain.
48035-8.1-27E
AID:114681 | 06/05/2018
a
ln
130
We have an initial information wherein it’s given that a bowl of porridge at 200o F (too
hot) is placed in a 70o F room. One minute later, the porridge has cooled to 180o F .We
need to find out , when will the temperature become 120o F
According to Newton’s law of cooling , if we introduce a hot object in a cool surrounding
or vice versa then the rate at which the object cools or warms(in the other case) is not
proportional to its temperature but rather to the difference in temperature between the
object and its surrounding.
We can use a formula here, y(t) Aekt T , where T the temperature of the
a
a
surrounding (the ambient temperature which we can consider to be constant),
y(t) temperature of the object at time t
Considering the equation
We have
y(t) Aekt T -------
(ii)
y(0) Aek*0 70 A 70 200 ,substituting the value as t 0
,since it’s the initial condition and Ta 70 (given in the problem) ----- (i)
From (i), we get A 200 70 130 -----
(iii)
A minute later, the condition becomes y(1) 130ek*1 70 180 , considering t 1
Therefore,
130
e
k
180
70
110
Applying natural logs on both the sides, ln(130) * k ln110
Thus
110
k
ln( ) ------
(iv)
130
We need to find out , when will the temperature become 120o F
Thus from(ii), we get, 120 130e ln(110 )*t
130
70 ,consideringy(t) 120 and
substituting the values from (iii)& (iv)
120 70 130e
ln(110 )*t
130
Therefore, 50 130e ln(110 )*t
130
Thus, 50
130
ln(110 )*t
e 130
Applying natural logs on both the sides
ln( 50 )
ln(110)*
t
130 130
50
130
t
ln 110
5.72min
The temperature would become 120o F after 5.72min
48035-8.1-28E
AID:114681 | 06/05/2018
a
We have an initial information wherein it’s given that a bowl of porridge at 200o F (too
hot) is placed in a 70o F room. One minute later, the porridge has cooled to 160o F .We
need to find out , when will the temperature after 5.72min
According to Newton’s law of cooling , if we introduce a hot object in a cool surrounding
or vice versa then the rate at which the object cools or warms(in the other case) is not
proportional to its temperature but rather to the difference in temperature between the
object and its surrounding.
We can use a formula here, y(t) Aekt T , where T the temperature of the
a
a
surrounding (the ambient temperature which we can consider to be constant),
y(t) temperature of the object at time t
Considering the equation
We have
y(t) Aekt T -------
(ii)
y(0) Aek*0 70 A 70 200 ,substituting the value as t 0
,since it’s the initial condition and Ta 70 (given in the problem) ----- (i)
From (i), we get A 200 70 130 -----
(iii)
A minute later, the condition becomes y(1) 130ek*1 70 160 , considering t 1
Therefore,
130
e
k
160
70
90
Applying natural logs on both the sides, ln(130) * k ln 90
Thus
k
ln( 90 ) ------
(iv)
130
We need to find out , when will the temperature after 5.72min
ln( 90 )*5.72
Thus from(ii), we get, y(5.72) 130e 130
70 ,considering t 5.72min and
substituting the values from (iii)& (iv)
y(5.72) 130e
(
0.3677)*5.72
70
Therefore,
Thus,
y(5.72) 130e2.1034 70
y
(5.72)
130*0.122
70
y
(5.72)
15.865
70
85.865
We need to find out , when will the temperature after 5.72minand the answer to that
question is 85.865o F
48035-8.1-29E
AID:114681 | 06/05/2018
a
We have an initial information wherein it’s given that a cold drink at 50o F is placed in a
70o F room. Two minute later, the cold drink has been warmed to 56o F .We need to find
out , what will be the temperature at time, t .
According to Newton’s law of cooling, if we introduce a hot object in a cool surrounding
or vice versa then the rate at which the object cools or warms(in the other case) is not
proportional to its temperature but rather to the difference in temperature between the
object and its surrounding.
We can use a formula here, y(t) Aekt T , where T the temperature of the
a
a
surrounding (the ambient temperature which we can consider to be constant),
y(t) temperature of the object at time t
Considering the equation
We have
y(t) Aekt T -------
(ii)
y(0) Aek*0 70 A 70 50 ,substituting the value as t 0
,since it’s the initial condition and Ta 70 (given in the problem) ----- (i)
From (i), we get A 50 70 20 -----
(iii)
Two minutes later, the condition becomes y(2) (20)ek*2 70 56 , considering t 2
Therefore,
Applying natural logs on both the sides,
Thus
20
e
2k
56
70
14
ln(20)* 2
k
ln14
k 1 ln(14 ) ------ (iv)
2 20
We need to find out , what will be the temperature at time, t .
1 ln
14
Thus from(ii), we get, y(t) 20e2 20 *t 70 ,substituting the values from (iii)& (iv)
1 ln
14
The temperature at time, t would become
y
(
t
)
20
e
2
20
*t
70
48035-8.1-30E
AID:114681 | 06/05/2018
a
20
20
We have an initial information wherein it’s given that a cold drink at 50o F is placed in a
70o F room. Two minute later, the cold drink has been warmed to 56o F .We need to find
out , what will be the temperature after10 min . We also need to find out, when the
temperature becomes 66o F
According to Newton’s law of cooling, if we introduce a hot object in a cool surrounding
or vice versa then the rate at which the object cools or warms(in the other case) is not
proportional to its temperature but rather to the difference in temperature between the
object and its surrounding.
We can use a formula here, y(t) Aekt T , where T the temperature of the
a
a
surrounding (the ambient temperature which we can consider to be constant),
y(t) temperature of the object at time t
Considering the equation
We have
y(t) Aekt T -------
(ii)
y(0) Aek*0 70 A 70 50 ,substituting the value as t 0
,since it’s the initial condition and Ta 70 (given in the problem) ----- (i)
From (i), we get A 50 70 20 -----
(iii)
Two minutes later, the condition becomes y(2) (20)ek*2 70 56 , considering t 2
Therefore,
Applying natural logs on both the sides,
Thus
20
e
2k
56
70
14
ln(20)* 2
k
ln14
k 1 ln(14 ) ------ (iv)
2 20
We need to find out , what will be the temperature after 10 min .
Thus from(ii), we get,
1 ln
14
y
(10)
20
e
2
*10
70
20
e
1.78
70
(
20* 0.1686)
70
3.37
70
66.627
,substituting the values from (iii)& (iv)
We also need to find out, when the temperature becomes 66o F
1 ln
14
66 20e2
*t
70 20e 0.178t 70
4 20e0.178t
From (ii), we get,
4 e0.178t
20
applying, ln
1 1
ln( ) t
0.178 5
1.609 t 9.04
0.178
The temperature after 10mins would become 66.627o F
After 9.04min , the temperature would become 66o F
48035-8.1-31E
AID:114681 | 06/05/2018
a
ln
10
We have an initial information wherein it’s given that the room temperature is 70o F . In
2min s that's between 10 : 07 pm &10 : 09 pm , the martini warms from 60o Fto61o F .We
need to note that the secret agent was found murdered at 10 : 07 pm The secret agent’s
martini is always served at 40o F .We need to find the time, t when the martini was at a
temperature which is 40o F
According to Newton’s law of cooling, if we introduce a hot object in a cool surrounding
or vice versa then the rate at which the object cools or warms(in the other case) is not
proportional to its temperature but rather to the difference in temperature between the
object and its surrounding.
We can use a formula here, y(t) Aekt T , where T the temperature of the
a
a
surrounding (the ambient temperature which we can consider to be constant),
y(t) temperature of the object at time t
Considering the equation
We have
y(t) Aekt T -------
(ii)
y(0) Aek*0 70 A 70 60 ,substituting the value as t 0
,since it’s the initial condition and Ta 70 (given in the problem) ----- (i)
From (i), we get A 60 70 10 -----
(iii)
Two minutes later, the condition becomes y(2) (10)ek*2 70 61 , considering t 2
Therefore,
Applying natural logs on both the sides,
10ek*2 61 70 9
ln(10) * 2
k
ln 9
9
10
Thus k ------ (iv)
2
We need to find the time, t when the martini was at a temperature which is 40o F
ln
9
1
Thus from(ii), we get,
(iv)
y
(
t
)
40
10
e
10
*t*2
70 ,substituting the values from (iii)&
ln
9
1
Solving this 40 70 10e 10 *t*2
ln
9
1
30 10e 10 *t*2
30
ln 9 1
10 e
10 *t*2
Applying natural logs on both the sides, we get,
ln 3 ln 9 *t * 1
t 2 ln 3 =
2.197
20.92
ln 9
10
1.05
So we can consider that
murder was done.
21min before the temperature of the martini was 60o F , the
Now, at 10:07pm the martini temperature was 60o F
Therefore, murder was done at 10 : 07 pm 21min 9 : 46 pm
The murder was done at 9:46pm
2
48035-8.1-32E
AID:114681 | 06/05/2018
a
92
92
a
We have an initial information wherein it’s given that the coffee’s temperature is 160o F ,
20min after it’s being served. At the 22nd min , the temperature is 158o F .Considering,
the room temperature as 68o F , we need to find out if the serving temperature is hotter or
cooler that 180o F .
According to Newton’s law of cooling, if we introduce a hot object in a cool surrounding
or vice versa then the rate at which the object cools or warms (in the other case) is not
proportional to its temperature but rather to the difference in temperature between the
object and its surrounding.
We can use a formula here, y(t) Aekt T , where T the temperature of the
a
a
surrounding (the ambient temperature which we can consider to be constant),
y(t) temperature of the object at time t
Considering the equation
We have
y(t) Aekt T -------
(ii)
y(20) Aek*20 68 160 -----
(iv)
,substituting the value as t 20,since it’s the initial condition and y(20) 160 ,
T 68o F (given in the problem)
92 Ae
20*k
Thus A 92
e20*k
------(iii)
Two minutes later, the condition becomes y(22) Aek*22 68 158 , considering t 22 ,
since it’s the initial condition and Ta 68 F , y(22) 158 (given in the problem)
o
Thus 90 Ae22k
A 90
------(i)
e22k
Therefore, we can see that (i)=(iii), Hence
92
e20k 90
e
22 k
e22k
e20k 90
92
e
22k
20k
90
92
e2k 90
92
Applying natural log on both the sides, we get,
2k ln 90
k 1 ln 90 0.01
2
Substituting the value of k 0.01in equation (iv), we get,
y(20) Ae
k*20
68 160
160 68 Ae
20*(
0.01)
92 Ae
0.2
92 A
e0.2
A 114.6
From (ii), we get,
y(t) Aekt T
In order to get the serving temperature, we need to consider
y(0) 114.6ek*0 68 114.6 68 182.6o F
Thus we can conclude that the serving temperature is greater that 180o F
a
48035-8.1-33E
AID:114681 | 06/05/2018
a
92
92
a
We have an initial information wherein it’s given that the coffee’s temperature is 160o F ,
20min after it’s being served. At the 22nd min , the temperature is 158o F .Considering,
the room temperature as 68o F , we need to find out if the serving temperature is hotter or
cooler that 180o F .
According to Newton’s law of cooling, if we introduce a hot object in a cool surrounding
or vice versa then the rate at which the object cools or warms (in the other case) is not
proportional to its temperature but rather to the difference in temperature between the
object and its surrounding.
We can use a formula here, y(t) Aekt T , where T the temperature of the
a
a
surrounding (the ambient temperature which we can consider to be constant),
y(t) temperature of the object at time t
Considering the equation
We have
y(t) Aekt T -------
(ii)
y(20) Aek*20 68 160 -----
(iv)
,substituting the value as t 20,since it’s the initial condition and y(20) 160 ,
T 68o F (given in the problem)
92 Ae
20*k
Thus A 92
e20*k
------(iii)
Two minutes later, the condition becomes y(22) Aek*22 68 158 , considering t 22 ,
since it’s the initial condition and Ta 68 F , y(22) 158 (given in the problem)
o
Thus 90 Ae22k
A 90
------(i)
e22k
Therefore, we can see that (i)=(iii), Hence
92
e20k 90
e
22 k
e22k
e20k 90
92
e
22k
20k
90
92
e2k 90
92
Applying natural log on both the sides, we get,
2k ln 90
k 1 ln 90 0.01
2
Substituting the value of k 0.01in equation (iv), we get,
y(20) Ae
k*20
68 160
160 68 Ae
20*(
0.01)
92 Ae
0.2
92 A
e0.2
A 114.6
From (ii), we get,
y(t) Aekt T
In order to get the serving temperature, we need to consider
y(0) 114.6ek*0 68 114.6 68 182.6o F
Thus we can conclude that the serving temperature is greater that 180o F
The serving temperature is 182.6o F
a
48035-8.1-34E
AID:114681 | 06/05/2018
a
110
a
We have an initial information wherein it’s given that the coffee’s temperature is 180o F ,
when being served. At the 2nd min , the temperature is165o F .Considering, the room
temperature as 70o F and the temperature after 5 mins as120o F , we need to find out the
serving temperature
According to Newton’s law of cooling, if we introduce a hot object in a cool surrounding
or vice versa then the rate at which the object cools or warms (in the other case) is not
proportional to its temperature but rather to the difference in temperature between the
object and its surrounding.
We can use a formula here, y(t) Aekt T , where T the temperature of the
a
a
surrounding (the ambient temperature which we can consider to be constant),
y(t) temperature of the object at time t
Considering the equation y(t) Aekt T -------
(ii)
Coffee’s temperature is 180o F , when being served.
y(0) Ae
k*0
T A T
a a
substituting, y(0) 180(given)
Thus,
T
70
o
F (given)
Now, we have
180 A 70
A 110
y(2) 110e
k*2
70 165
95 110e
2*k
------(iv)
95
110 e2*k
Applying natural logs on both the sides, we get,
2k ln 95
k 1 ln 95 0.07
2
110
The current status of temperature after 5 mins is 147.5o F
y(5) 110e
0.07*5
70
considering, k 0.07
y(5) 147.52
Now, we can see that the required temperature is 120o F
All the components including k,t,Ta , required temperature are constant and the only
component that can be changed is
temperature
Thus
y(5) Ae
0.07*5
70 120
considering, k 0.07
50 Ae0.07*5
A, and that’s how we can find the new serving
Thus, A 50
e5*0.07
70.95
o
F
Hence, y(0) 70.95 70 140.95o F
Thus, the serving temperature is 140.95o F
48035-8.1-35E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that the initial invest is $1000 at an
annual interest rate of 8% . We need to find out the value of the investment after 1 year
under the following forms of compounding: annual, monthly, daily, continuous.
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Annual compounding
Monthly compounding
1000(1
0.08)
1
1000 *1.08
1080
1
12
1000(1 0.08)1*12 1000 12.08
1000
1.006667
12
1000
*1.083
1083
12
12
Daily compounding =
1*365 365
0.08
365.08
365
1000
1
1000
1000(1.00021918)
1000*1.083278
1083.28
365
365
Continuous =1000e0.08*1 1000*1.083287 1083.29
48035-8.1-36E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that the initial invest is $1000 at an
annual interest rate of 8% . We need to find out the value of the investment after 1 year
under the following forms of compounding: annual, monthly, daily, continuous.
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Annual compounding
Monthly compounding
1000(1
0.08)
5*1
1000*1.08
5
1000*1.469328
1469.328
1
5*12
1000(1 0.08)5*12 1000 12.08
1000
1.0066667
60
1000*1.4898
1489.848668
12
12
Daily compounding =
5*365 365*5
0.08
365.08
365*5
1000
1
1000
1000(1.00021918)
1000*1.4917645336
1491.7645336
365
365
Continuous =1000e0.08*5 1000*1.49182469764 1491.82469764
48035-8.1-37E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that the initial invest is $10000at an
annual interest rate of12% , by a person named A in the year 1990.Again, there’s another
person named B who’s initial investment is the year 2000 is $20000at a rate of interest
12% We need to find out the value of the investment made the person named A and B in
2010
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Continuous compounding by person named A and person named B respectively are
10000e0.12*20 10000e2.4 110231.763806
20000e0.12*10 20000e1.2 66402.3384547
The value of investment made the person named A and B in 2010are
$110231.764&$66402.338
48035-8.1-38E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that the initial invest is $10000at an
annual interest rate of 4% , by a person named A in the year 1990.Again, there’s another
person named B who’s initial investment is the year 2000 is $20000at a rate of interest
4% We need to find out at what rate of interest, person named A ends up exactly even
with person named B. In this way, we need to find out, at what rate of interest person
named A, ends up getting $20000in the year 2000 which is exactly 10 years after his
initial investment
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Continuous compounding by person named A and person named B respectively are
10000e0.04*20 10000e0.8 22255.409
20000e0.04*10 20000e0.4 29836.49
Continuous compounding by person named A in 10 years
10000
e
r*10
20000
e
r*10
20000
2
10000
Applying natural logs on both the sides, we get
10r ln 2
r ln 2 0.0693
10
Thus the rate of interest, person A needs to get, in order to make his investment value
$20000by the year 2000 is 0.0693*100% 6.93%
The value of investment made the person named A and B in 2010are
22255.409 & 29836.49 at a rate of interest 4%
The rate of interest, person A needs to get, in order to make his investment value $20000
by the year 2000 is 6.93%
48035-8.1-39E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that the initial investment is $34 in the
year 1985 on a set of basketball card. Again, there’s a change in the book price by the
year 1995(10years later) and now it becomes $9800.We need to find out an equation for
the worth of the set at time t years, considering a constant percentage return in this
investment. We need to find out another thing, which is the value of the book in 2005, at
the same rate of return
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Exponential rise(+)/Exponential depreciation(-)=
v(t) value at time, t , r rate of interest
Linear depreciation v(t) mt b
v(t) Ae(rise/ depreciation)*r*t where
Now, in the year 1985, the value was $34
10 years later, in the year 1995, the value becomes $9800
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Here P principle 34, r rate of interest, t time 10, n 1 number of times
compounded annually
34(1 r)
10
9800
(1 r)
10
9800
34
1
1 r 9800
10
34
Thus,
1
r 9800
10
1
34
1
r (288.235294)10 1
r 0.762
Therefore, in the year 2005 that is, after 20 years, the value becomes
34(1.762)20 2824705.88 , we get this from the basic formula
Exponential rise(+)/Exponential depreciation(-)=
v(t) value at time, t , r rate of interest
v(t) Ae(rise/ depreciation)*r*t where
We have to consider exponential increase here, so v(t) Aer*t
v(t) value at time, t , r rate of interest
v
(0)
34
Ae
r*0
A
------(i)where
Initially
A
34
Thereby,
v(10) 9800 34e
10*r
9800 e10*r
34
, here
A
34,
t
10,
v
(10)
9800
Applying natural logs on both the sides, we get
r
1 ln(9800)
10 34
We need to find out an equation for the worth of the set at time t years, considering a
constant percentage return in this investment.
Thus , comparing with(i), we get,
1 ln( 9800 )t
v(t) 34e10 34
In the year 2005 that is, after 20 years, the value becomes $2824705.88.
The equation for the worth of the set at time t years, considering a constant percentage
1 ln( 9800 )t
return in this investment is v(t) 34e10 34
48035-8.1-40E
AID:114681 | 06/05/2018
22
We have an initial information wherein it’s given that the initial investment is $22 in the
year 1985 on a set of basketball card. Again, there’s a change in the book price by the
year 1995(10years later) and now it becomes $32 .We need to find out value of the book
in 2005 with the same rate of return
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Exponential rise(+)/Exponential depreciation(-)=
v(t) value at time, t , r rate of interest
Linear depreciation v(t) mt b
v(t) Ae(rise/ depreciation)*r*t where
Now, in the year 1985, the value was $22
10 years later, in the year 1995, the value becomes $32
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Here P principle 22, r rate of interest, t time 10, n 1 number of times
compounded annually
22(1 r)
10
32
(1 r)
10
32
22
1
1 r 32
10
22
Thus,
1
r 32
10
1
1
r (1.45)10 1
r 0.038
Therefore, in the year 2005 that is, after 20 years, the value becomes 22(1.038)20 46.545
, we get this from the basic formula
In the year 2005 that is, after 20 years, the value becomes $46.545.
48035-8.1-60E
AID:114681 | 06/05/2018
2 t
t
We have an initial information wherein it’s given that the value of a piece of land t years
from now is $40, 000e2 t . We need to find the time, t . It is also given that the present
value is 40, 000e2 t 0.06t and the annual inflation rate is 6%
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Exponential rise(+)/Exponential depreciation(-)=
v(t) value at time, t , r rate of interest
Linear depreciation v(t) mt b
Present value= 40, 000e2 t 0.06t
v(t) Ae(rise/ depreciation)*r*t where
Thus,
P(t) 40, 000e
2
t
0.06t
Differentiation, applied
dP 40, 000e
2
dt
t
0.06t
(
2
0.06)
dP 0, for max P(t)
dt
(
2
0.06) 0
2 t
2 0.06
2
1
0.06
t 277.78
Thus, t 277.78, would maximize the present value
t
48035-8.1-61E
AID:114681 | 06/05/2018
3
We have an initial information wherein it’s given that a business has an income stream of
$P(t). The present value at interest rate, r of this income for the next T years is
T
P(t)ertdt . We need to compare the present values at 5% for three people with the
0
following salaries for 3 years.
A : P(t) 60, 000
B : P(t) 60, 000 3000t
C : P(t) 60, 000e
0.05t
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Exponential rise(+)/Exponential depreciation(-)=
v(t) value at time, t , r rate of interest
Linear depreciation v(t) mt b
v(t) Ae(rise/ depreciation)*r*t where
Present values for A:
3
P
(
t
)
60,
000
e
0.05t
dt
60,
000
e
0.05t
3
1200000*
e
0.15
60,
000
167150.4283
.05
0
0.05
Present values for B can be calculated from below
P(t)
60,
000e
0.05t
dt
60,
000
e
0.05t
3
1200000
*
e
0.15
60,
000
167150.4283
(i)
.05 0 0.05
3 3 3
B : P(t)
(60, 000 3000t)e
0.05t
dt
60, 000e
0.05t
dt
3000te
0.05t
0 0 0
te0.05t e0.05t 3
e
0.05t
3
From(i) 167150.4283 3000[
]3 167150.4283 3000 51.6425
0.05 0 0.05
0.05
2
167150.4283 3000
4.0743
179373.3566
0
Present values for C:
C : P(t) 60, 000e
0.05t
3 3
P(t)e0.05tdt 60, 000dt 180000
0 0
Hence we have A<B<C
Present values for A,B,C are $167150.4283, $179373.3566, $180000
0
0
48035-8.1-62E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that the future value of an income
T
stream after T years at interest rate, r is given by P(t)ertdt . We need to find the future
0
value for cases A,B and C. We need to use the below value
3
A : P(t)
60, 000e
0.05t
dt
0
3
B : P(t)
(60, 000 3000t)e
0.05t
dt
0
C : P(t) 60, 000e
0.05t
We need to find the difference between what present value and future value measure.
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Exponential rise(+)/Exponential depreciation(-)=
v(t) value at time, t , r rate of interest
Linear depreciation v(t) mt b
v(t) Ae(rise/ depreciation)*r*t where
Future values for A:
3 3 3
0.05(3t ) 0.15 0.05t 0.05t
P(t) 60, 000e dt 60, 000e e dt 69710.05 e dt 0 194201.8913
0 0
Present values for B can be calculated from below
3 3
P(t) (60, 000 3000t)e0.05(3t)dt e0.15 (60, 000 3000t)e0.05tdt 208402.1079
0 0
Present values for C:
C : P(t) 180000e0.15t 209130.1637
Future values for A,B,C are $194201.8913, $208402.1079, $209130.1637
48035-8.1-63E
AID:114681 | 06/05/2018
We have an initial information where it’s given that $280,000 is given in 4 installments.
So we need to find out that accepting $280,000 in 4 installments is better than getting
$1,000,000. Note that 8% interest is involved in both and that the payments are made at
the beginning of the year.
If we let y(t) represent the number of bacteria in a culture at time t, then the rate of
change of population with respect to time is
y(t) , we have
y' (t) . Thus, since y' (t) is proportional to
y' (t) k y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
According to the problem $280,000 is given out in 4 installments and hence we get the
below equation
(280, 000 *1.083 ) (280, 000 *1.082 ) (280, 000 *1.08) 280, 000
280, 000 *1.08(11.08 1.08
2
) 280, 000
302, 400(11.08 1.1664) 280, 000
1, 261, 711.36
Now, considering $1,000,000, given out with an interest rate as 8%; and the final amount
is: 1, 000, 000e0.08*3 1, 000, 000e0.24 1, 000, 000*1.27 1, 271, 249.15
Thus, we can understand that getting $1,000,000 at an interest of 8% is way better than
getting $280,000 in 4 installments with 8% rate of interest.
48035-8.1-64E
AID:114681 | 06/05/2018
We have an initial information where the rate is given which is 8%. We need to find out
why rule of 72 is still used, though the rule of 69 is better.
If we let y(t) represent the number of bacteria in a culture at time t, then the rate of
change of population with respect to time is
y(t) , we have
y' (t) . Thus, since y' (t) is proportional to
y' (t) k y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
Let’s consider a value , 1000
Considering the equation
We have
y(t) Aekt ------ (i)
y
(0)
1000
Based on (i), we have
y
(0)
1000
=
Aek*0 = A (since e0 1 )
Therefore, 1000= A
Now, we need to find the doubling time
Therefore, 2 y(0) 2 *1000 2000
y(t) 2000 1000e
0.08t
2000 e0.08t
1000
2 e0.08t
Thus 1
0.08 ln 2 t
0.693 t 8.66
0.08
t 8.66
Now applying rule of 72= 72 9 which is a whole number
8
Applying rule of 69, we get 69 8.625
8
This rule of 72 is used because it gives whole number result which is easier to find out,
though rule of 69 gives a very close result
The doubling time, t 8.66
48035-8.1-41E
AID:114681 | 06/05/2018
28
28
We have an initial information wherein it’s given that in 1975, income between
$16000&$20000 was taxed at 28%. In 1988, that is after 13 years, income between
$16000&$20000 was taxed at 15%. We need to prove that taking inflation into account,
we need to prove that this is not a valid comparison.
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Exponential rise(+)/Exponential depreciation(-)=
v(t) value at time, t , r rate of interest
Linear depreciation v(t) mt b
v(t) Ae(rise/ depreciation)*r*t where
Now in year 1975, the tax was 28%, so we can consider that the initial condition
v
(0)
28 ------
(i)
Exponential depreciation= v(t) Aer*t
v(0) 28 Aek*0
From the equation (i), we get
A
28
In the year 1988, the tax was 15% which was after 13 years from 1975
y(13) 15 28e
k*13
15 ek*13
28
Applying natural logs on both the sides, we get,
ln 15 k *13
1 ln 15 k
13
Now, let’s consider that someone bought an asset worth $16000in the year 1975. So this
can be considered a depreciating asset. The reason is that, after years the asset value
depreciates and also does the currency value in a similar way.
Thus, let’s consider the initial condition
v(0) 16000 Ae
k*0
A
A 16000
v(13) 16000e
k*13
considering, k 1 ln(15 )
13 28
1
15
ln *13
v(13) 16000e
13
28
v(13) 16000 * 15 $8571.43
28
Considering $8571.43 as the value same $16000, 13 years later; we can see that 28% tax
would give us a tax amount of 8571.43 * 28 $2400.0004 ------- (i)
100
13 years later, in the year 1988, the tax value of $16000 at the rate of 15% becomes
16000 *15 $2400 = equation (i)
100
Thus, we can conclude that inflation is the only criterion which needs to be considered
here; and that taxes are reduced based on the inflation and there’s never a real decrease in
the tax amount that we owe to the government.
48035-8.1-42E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that in 1988, income above $30000 was
taxed at 28%. Average inflation rate was 5.5% between 1975 and 1988 which 13 years.
Similarly, from 1975, $16000 faced an inflation of an average rate of 5.5% for 13 years.
We need to find the tax percent for the year 1975.
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Exponential rise(+)/Exponential depreciation(-)=
v(t) value at time, t , r rate of interest
Linear depreciation v(t) mt b
v(t) Ae(rise/ depreciation)*r*t where
Now in year 1988, the tax was 28%, so we can consider that the initial condition
v
(0)
28 ------
(i)
Exponential depreciation= v(t) Aer*t
v(0) 28 Aek*0
From the equation (i), we get
A
28
From the year 1975, year after year, inflation at the rate of 5.5% was initiated and the tax
rate became
y(13) 28e0.055*13 57.24%
(ii)
Thus, let’s consider the initial condition
v(0) 16000 Ae
k*0
A
A 16000
v(13) 16000e
0.055*13
7827.07
considering, k 5.5 / 100 0.055
Considering $7827.07 as the value same $16000, 13 years later; we can see that 57.24%
tax would give us a tax amount of 7827.07 *57.24 $4480.21487 ------- (i)
100
13 years later, in the year 1988, the tax value of $16000 at the rate of 28% becomes
16000 * 28 $4480 = equation (i)
100
Based on equation(ii), the tax rate in the year 1975 was 57.24%
48035-8.1-43E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that the first $30,000 is taxed at 15%,
the remainder is taxed at 28%. We need to find the tax amount for $40000. We need to
find the tax amount for $42000.
Let’s consider that there’s an inflation of 5% in an year and the living cost becomes
$42000. We need to find the tax amount for $42000 based on the previous tax rates
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Exponential rise(+)/Exponential depreciation(-)=
v(t) value at time, t , r rate of interest
Linear depreciation v(t) mt b
v(t) Ae(rise/ depreciation)*r*t where
We have an initial information wherein it’s given that the first $30,000 is taxed at 15%
So the tax amount is $30000 * 15 $4500
100
Above $30000, the tax is 28% and so the tax on remaining $10000 is
$10000 * 28
100
$2800
Above $30000, the tax is 28% and so the tax on remaining $12000 for $42000 is
$12000 * 28
100
$3360
The total tax is $4500 $2800 $7300on $40000
The total tax is $4500 $3360 $7860on $42000
Now, considering an inflation of 5% on previous tax amount, we have a new tax value
which is $7300 *105 $7665
100
The total tax on $40000, before was $7300
The total tax on $42000, before was $7860
The total tax on $42000, now is $7665
48035-8.1-44E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that the first $30,000 is taxed at 15%,
the remainder is taxed at 28%. We need to find the tax amount for $40000. We need to
find the tax amount for $42000.
Let’s consider that there’s an inflation of 5% in an year and the living cost becomes
$42000. We need to find the tax amount for $42000 based on the previous tax rates
Now, we would finally find out that in one year the value of $42000 depreciates to
$40000 at a rate 5%
We need to make the tax owed previously, same as the current tax owed.
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Exponential rise(+)/Exponential depreciation(-)=
v(t) value at time, t , r rate of interest
Linear depreciation v(t) mt b
v(t) Ae(rise/ depreciation)*r*t where
We have an initial information wherein it’s given that the first $30,000 is taxed at 15%
So the tax amount is $30000 * 15 $4500
100
Above $30000, the tax is 28% and so the tax on remaining $10000 is
$10000 * 28
100
$2800 -----
(i)
Above $30000, the tax is 28% and so the tax on remaining $12000 for $42000 is
$12000 * 28
100
$3360
The total tax is $4500 $2800 $7300on $40000
The total tax is $4500 $3360 $7860on $42000
Now, considering an inflation of 5% on previous tax amount, we have a new tax value
which is $7300 *105 $7665
100
We need to make the tax owed previously, same as the current tax owed.
From (i), we get that the tax slot after $30000 is 28% and we need to change this rate of
interest to get the same value that is $2800
Let the new rate be r
Thus, comparing with (i), we get,
$12000* r
100 $2800
r 2800 *100 23.3%
12000
We need to make the tax owed previously, same as the current tax owed; and for that we
need to change the tax to 23.3% for $30000 and above
48035-8.1-45E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that $40,000 asset decreases at constant
percentage rate of 10%. We need to find out the asset worth after 10 years, 20 years.
Also, we need to find out the asset worth after 10 years, considering linear depreciation
and there’s another condition wherein under this particular linear depreciation; the asset
value becomes $0 in 20 years.
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Exponential rise(+)/Exponential depreciation(-)=
v(t) value at time, t , r rate of interest
Linear depreciation v(t) mt b
v(t) Ae(rise/ depreciation)*r*t where
Exponential depreciation(-)= v(t) Aer*t
v(0) 40000 Ae
r*0
A 40, 000
,considering initial value as $40,000
10 years later the asset value becomes
v(10) Ae
0.1*10
$14715.18
considering, A 40, 000
,depreciation rate r 10% 0.1
20 years later the asset value becomes
v(20) Ae
0.1*20
$5413.41
considering, A 40, 000
Linear depreciation v(t) mt b
Under this particular linear depreciation; the asset value becomes $0 in 20 years.
v(20) 0 m * 20 40000
Thus m 40000 2000
20
considering initial value as $40,000
The value of the asset, after 10 years, considering linear depreciation is
v(10) (2000)*10 40000 20000
considering, m 40000 2000
20
Under exponential depreciation, the asset value after 10 and 20 years are $14715.18&
$5413.41respectively
Under linear depreciation, the asset value after 10 and 20 years are $20000& 0
respectively
48035-8.1-46E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that $40,000 asset decreases at constant
percentage rate of 40%. We need to find out the asset worth after 5 years, 10 years. Also,
we need to find out the asset worth after 5 years, considering linear depreciation and
there’s another condition wherein under this particular linear depreciation; the asset value
becomes $0 in 10 years.
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Exponential rise(+)/Exponential depreciation(-)=
v(t) value at time, t , r rate of interest
Linear depreciation v(t) mt b
v(t) Ae(rise/ depreciation)*r*t where
Exponential depreciation(-)= v(t) Aer*t
v(0) 40000 Ae
r*0
A 40, 000
,considering initial value as $40,000
5 years later the asset value becomes
v(5) Ae
0.4*5
$5413.41
considering, A 40, 000
,depreciation rate r 40% 0.4
10 years later the asset value becomes
v(10) Ae
0.4*10
$732.63
considering, A 40, 000
Linear depreciation v(t) mt b
Under this particular linear depreciation; the asset value becomes $0 in 20 years.
v(10) 0 m *10 40000
Thus m 40000 4000
10
considering initial value as $40,000
The value of the asset, after 05 years, considering linear depreciation is
v(5) (4000)*5 40000 20000
considering, m 40000 4000
10
Under exponential depreciation, the asset value after 5 and 10 years are $5413.41&
$732.63 respectively
Under linear depreciation, the asset value after 5 and 10 years are $20000& 0
respectively
48035-8.1-47E
AID:114681 | 06/05/2018
16
16
16
1. We need to prove that the slopes for population/time graph from 1000 to 500 is
different than that of population/time graph from 10 to 5
2. Again we need to prove that slopes for population/time graph from ln1000 to ln500 is
exactly the same as that of population/time graph from ln10 to ln5
3. We need to find the graph of population vs time, considering an increase with constant
percentage rate.
4. Finally, under the same condition which is increase in constant percentage rate, we
need to see how the graph looks like for logarithm of population vs time
y
y
According to the basic formulas: slope 2 1 =change in y axis/ change in x axis
x
x
2 1
Let’s consider a table
hours
Population
0
1600
0.68
1000
1
800
1.6857
500
2
400
3
200
4
100
We are initiating a population of 1600 at 0 hour. We are reducing the population to half,
every hour.
We need to find the slope for 1000 to 500 population drop
Hence in the above table we have found the time associated to population of 1000 and
500
formula : y(t) Ae
kt
Table : y(0) 1600 Ae
k*0
A;thus, A 1600
Table : y(4) 100 1600e
4k
; k rate
1 ln 1 k 0.69
4
Pr oblem : 500 1600e
0.69*t
1 ln 5 t 1.6857hrs
0.69
Pr oblem :1000 1600e
0.69*t
1
ln 10 t 0.68hr
0.69
Slope for 1000 to 500 drop
1000
500
500
497.166 -------
(i)
0.68
1.6857
1.0057
Let’s plot the population vs time graph for this table
Again this consider the below table
hours
Population
0
16
0.68
10
1
8
1.6857
5
2
4
3
2
4
1
With the above table, we get the following graph
Slope for 10 to 5 drop 10 5
5
4.97166 -----
(ii)
0.68
1.6857
1.0057
Comparing (i) and (ii), we get that the slopes from the drops are different
Let’s take another table
hours
Population
ln(population)
0
1600
7.3777589
0.68
1000
6.907755
1
800
6.6846
1.6857
500
6.2146
2
400
5.99
3
200
5.298
4
100
4.605
Thus we get
hours
ln(population)
0
7.3777589
0.68
6.907755
1
6.6846
1.6857
6.2146
2
5.99
3
5.298
4
4.605
Thus from the above table information, we get the below graph
Slope for ln(1000) to ln(500) drop
6.907755
6.2146
0.693155
0.689 ------
(iii)
We need to consider another table
0.68
1.6857
1.0057
hours
Population
ln(population)
0
16
2.7725887
0.68
10
2.302585
1
8
2.079
1.6857
5
1.609
2
4
1.38629
3
2
0.69
4
1
0
Thus, we get the below table
hours
ln(population)
0
2.7725887
0.68
2.302585
1
2.079
1.6857
1.609
2
1.38629
3
0.69
4
0
Thus from the above table information, we get the below graph
Slope for ln(10) to ln(5) drop
2.302585
1.609 0.693585
0.689 -----
(iv)
Thus we can see that (iii)=(iv)
0.68
1.6857
1.0057
We need to find the graph of population vs time, considering an increase with constant
percentage rate.
Thus table would like
hours
Population
0
1
1
2
2
4
3
8
4
16
Thus the graph of the above table would look like
Again from the below table, we get
hours
Population
ln(population)
0
1
0
1
2
0.69
2
4
1.38629
3
8
2.079
4
16
2.7725887
So we get the following table from the above table
hours
ln(population)
0
0
1
0.69
2
1.38629
3
2.079
4
2.7725887
The graph is applicable for the above table
48035-8.1-48E
AID:114681 | 06/05/2018
It has been conjected that half the people who have ever lived today are still today. To see
whether this is plausible, we have to assume that humans have maintained a constant
birthrate b and death rate d . We need to prove that under the condition b 2d , half the
people are still alive today.
Let’s consider that 1000 people ever lived
So let’s consider the condition b 2d
So let’s consider that the surviving people count , as of today,is 500
b 2d
According to the condition Thus, b d , as of today
2
According to today’s condition, the population is
b
500 d
2
Now the above condition needs to be twice the condition of people who ever lived
b b
1000 d 2 *(500 d )
2 2
b
1000 d 1000 b 2d
2
Thus b
d
2
b
Thus, d
2
Hence it is proved that under the condition b 2d , half the people are still alive today.
48035-8.1-49E
AID:114681 | 06/05/2018
We have an initial population which is 100. Based on this initial population, after 4 hours,
the new population is 100*2 200 . Thus we need to plot a table in order to get a graph
for ln(population)vs time.
Later over the graph, on the linear line; we need to take 2 points; draw a line through
them and find the subsequent equation.
Finally, we need to find the population function
p(x) e
y(x)
If we let y(x) represent the number of bacteria in a culture at time t, then the rate of
change of population with respect to time is y' (x) . Thus, since
have y(x)
y' (x) is proportional to, we
y' (x) ky(x)
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(x) Aekx , where A=constant, k=growth constant x=time
Considering the equation
We have
y(x) Aekx ------ (i)
y
(0)
100
Based on (i), we have y(0) 100 = Aek*0 = A (since e0 1 )
Therefore, 100= A
Subsequently, we can consider that, doubling of the population happens every 4 hours
Now, let’s make a table out of the concept that, initial population is 100 and doubling of
the population happens every 4 hours
Time,x
Population,y(x)
Logarithm of population
0
100
4.6
4
200
5.298
8
400
5.99
12
800
6.6846
16
1600
7.378
20
3200
8.07
24
6400
8.76
Consolidating our data
Time(x)
Logarithm of population (y)
0
4.6
4
5.298
8
5.99
12
6.6846
16
7.378
20
8.07
24
8.76
Consecutive graph is
We know that slope of a line y mx c ,m=change in y/change in x
4.6 5.298 0.698
m 0.1745
0 4 4
5.298 0.1745* 4 c
5.298 0.698 c
c 4.6
Thus the equation of the line is
y
0.1745
x
4.6
Therefore the population function, p(x) e
y(x)
e
0.1745x
4.6
The equation of the line is
y
0.1745
x
4.6
The population function, p(x) e
y(x)
e
0.1745x
4.6
48035-8.1-50E
AID:114681 | 06/05/2018
We have an initial population which is 100. Based on this initial population, after 4 hours,
the new population is 100*2 200 .
We need to compare our regression model equation to the population model equation
If we let y(x) represent the number of bacteria in a culture at time t, then the rate of
change of population with respect to time is y' (x) . Thus, since
have y(x)
y' (x) is proportional to, we
y' (x) ky(x)
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(x) Aekx , where A=constant, k=growth constant x=time
Considering the equation
We have
y(x) Aekx ------ (i)
y
(0)
100
Based on (i), we have y(0) 100 = Aek*0 = A (since e0 1 )
Therefore, 100= A
Subsequent
Now we get
y(4) 200 100ek*4 , based on the problem; replacing
x=4,y(x)=200, A=100, x=4
2 e4k ,solving the above equation
Applying natural logs on both sides, ln 2 4k
Thus k ln 2 0.173
4
Subsequently, we can consider that, doubling of the population happens every 4 hours
Now, let’s make a table out of the concept that, initial population is 100 and doubling of
the population happens every 4 hours
Time,x
Population,y(x)
Logarithm of population
0
100
4.6
4
200
5.298
8
400
5.99
12
800
6.6846
16
1600
7.378
20
3200
8.07
24
6400
8.76
Again for x=8 and P(x)=400, we get
y(8) 400 100e
k*8
4 ek*8
k ln 4 0.173
8
Hence we can conclude that the value of k remains the same throughout
Now, let’s consider the equation p(x) ey(x)
In here, y(x) is supposed to be the equation of a straight line
Therefore,
y mx c
change( y axis)
4.6 5.298
m 0.1745
change(x axis) (0 4)
Again, 5.2 0.1745* 4 c
Hence, c 4.502
Equation : y 0.1745x 4.502
Population, P(x) e
0.1745x
4.502
Now, based on our generalized equation, where k=m(straight line) & A=c(straight line)
Population, P(x) e0.173x4.6
Copied from https://planetcalc.com/5992/
Linear regression
y=233.0357x+−982.1429y=233.0357x+−982.1429
Linear correlation coefficient
0.8779
Coefficient of determination
0.7707
Average relative error, %
224.15
Results
i x y Linear regression
i
x
y
Linear
regression
1
0
100
-982.1429
2
4
200
-50
3
8
400
882.1429
4
12
800
1814.2857
5
16
1600
2746.4286
6
20
3200
3678.5714
7
24
6400
4610.7143
Linear regression
Equation:
a coefficient
b coefficient
Linear correlation coefficient
Coefficient of determination
Standard error of the regression
Quadratic regression
Equation:
System of equations to find a, b and c
Copied from https://planetcalc.com/5992/
Exponential regression
y=e4.6052+0.1733xy=e4.6052+0.1733x
Correlation coefficient
1
Coefficient of determination
1
Average relative error, %
0.00
Results
i
x
y
Exponential
regression
1
0
100
100.0000
2
4
200
200.0000
3
8
400
400.0000
4
12
800
800.0000
5
16
1600
1600.0000
6
20
3200
3200.0000
7
24
6400
6400.0000
Exponential regression
Equation:
b coefficient
a coefficient
Correlation coefficient, coefficient of determination, standard error of the
regression - the same as above.
Please note that the exponential regression is almost similar to 49E
The equation of the line is y 0.1745x 4.502 ,thereby it’s corresponding regression
model based on population equation becomes P(x) e0.1745x4.502
The population function, based on just the exponential population is
p(x) ey( x) e0.173x4.6
48035-8.1-51E
AID:114681 | 06/05/2018
We have to consider the equation p(x) ey(x) . We need to find an exponential model for
the population data (0,10), (1,15), (2, 22),(3, 33),(4, 49) . We need to consider time,
x
,
time
;
population
,
p
(
x
); ln(
population
,
y
(
x
))
If we let y(t) represent the number of bacteria in a culture at time t, then the rate of
change of population with respect to time is
y(t) , we have
y' (t) . Thus, since y' (t) is proportional to
y' (t) k y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
We need to make a table now with some information
Time(x)
Population,p(x)
y(x)
0
10
2.3
1
15
2.7
2
22
3.09
3
33
3.49
4
49
3.89
Considering the equation,
From the chart, we get,
y(2) 22 10e
2k
22 e2k
10
ln(2.2) 2k
k 1 ln(2.2) 0.394
2
y(t) Aekt
y(0) 10, y(2) 22 10e2k
Considering the information, k 0.394 ,and
Thus, we get the equation, p(x) e
0.394x
2.303
y
(0)
2.303
48035-8.1-52E
AID:114681 | 06/05/2018
We have to consider the equation p(x) ey(x) . We need to find an exponential model for
the population data (0, 20), (1,16), (2,13),(3,11), (4, 9) . We need to consider time,
x
,
time
;
population
,
p
(
x
); ln(
population
,
y
(
x
))
If we let y(t) represent the number of bacteria in a culture at time t, then the rate of
change of population with respect to time is
y(t) , we have
y' (t) . Thus, since y' (t) is proportional to
y' (t) k y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
We need to make a table now with some information
Time(x)
Population,p(x)
y(x)
0
20
2.9957
1
16
2.77
2
13
2.56
3
11
2.397895
4
9
2.1972
Considering the equation,
From the chart, we get,
y(2) 16 20e
2k
16 e2k
20
ln(0.8) 2k
k 1 ln(0.8) 0.112
2
y(t) Aekt
y(0) 20, y(2) 16 20e2k
Considering the information, k 0.112 ,and
Thus, we get the equation, p(x) e
0.112x
2.9957
y
(0)
2.9957
48035-8.1-53E
AID:114681 | 06/05/2018
We have to consider the equation p(x) ey(x) . We need to find an exponential model for
1
the population data.
year
1960
1970
1980
1990
% population
farm
7.5
5.2
2.5
1.6
We need to consider time, x, time; population, p(x); ln( population, y(x))
If we let y(t) represent the number of bacteria in a culture at time t, then the rate of
change of population with respect to time is
y(t) , we have
y' (t) . Thus, since y' (t) is proportional to
y' (t) k y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
We need to make a table now with some information
Time(x)
Population,p(x)
y(x)
0
7.5%
2.015
10
5.2%
1.65
20
2.5%
0.92
30
1.6%
0.47
Considering the equation,
From the chart, we get,
y(10) 5.2 7.5e
10k
5.2 e10k
7.5
ln(0.693) 10k
y(t) Aekt
y(0) 7.5, y(10) 5.2 7.5e10k
k 1 ln(0.693) 0.0366
10
Considering the information, k 0.0366 ,and y(0) 7.5
From the chart, we get,
y(20) 2.5 7.5e10k
2.5 e20k
7.5
ln( ) 20k
3
k 1 ln(1) 0.055
20 3
y
(0)
7.5
ln(7.5)
2.015
y(20) 2.5 7.5e10k
Thus, we get the equation, p(t) e
0.055t
2.015
The required equation is, p(t) e
0.055t
2.015
48035-8.1-54E
AID:114681 | 06/05/2018
We have to consider the equation p(x) ey(x) . We need to find an exponential model for
the population data.
year
1960
1970
1980
1990
% population
farm
69.9
73.5
73.7
75.2
We need to consider time, x, time; population, p(x); ln( population, y(x))
If we let y(t) represent the number of bacteria in a culture at time t, then the rate of
change of population with respect to time is
y(t) , we have
y' (t) . Thus, since y' (t) is proportional to
y' (t) k y(t )
For some constant of proportionality k(the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k=growth constant t=time
We need to make a table now with some information
Time(x)
Population,p(x)
y(x)
0
69.9%
4.247
10
73.5%
4.297
20
73.7%
4.3
30
75.2%
4.32
Considering the equation,
From the chart, we get,
y(20) 73.7 69.9e
20k
1 ln(1.054) k
20
1 * 0.0529 k
20
k 0.0026
y
(0)
69.9
ln(69.9)
4.247
y(t) Aekt
y(20) 73.7 69.9e20k
Thus, we get the equation, p(t) e
0.0026t
4.247
The required equation is, p(t) e
0.055t
2.015
48035-8.1-55E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that half-life of amitriptyline is
31 46hours . We need to find out the quantity that would remain after 24hours,
considering 150gms of amitriptyline are in the blood stream of a person for whom the
half-line is 31 hours. We also need to find out the quantity that would remain after
24hours, considering 150gms of amitriptyline are in the blood stream of a person for
whom the half-line is 46 hours. We need to compare these conditions.
If we let y(t) represent the ‘some quantity’ at time t, then the rate of change of ‘some
quantity’ with respect to time is y' (t) . Thus, since y' (t) is proportional to y(t) , we have
y' (t) k y(t )
For some constant of proportionality k (the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k =growth constant t=time
Considering the equation
We have
y(t) Aekt ------ (ii)
y(0) Aek*0 A 150(given) ------
(i)
The value of the growth constant, considering half-life as 31 hours is
1 y(0) 1 A 150ek*31 , we get the values from (i)and (ii)
2 2
Therefore 1 *150 ek*31 , half-life amount is half of the initial amount
2 150
Thus, applying natural logs on both sides, we get
ln 1 31* k , now we know that ln 1 ln 2
Finally
2 2
k ln 2 0.0224 ------
(iii)
31
Thus the amount of amitriptyline that is present in the blood stream of a person for whom
the half-line is 31 hours , after 24 hours is = y(24) 150e0.0224*24 87.62
Again, the value of the growth constant, considering half-life as 46 hours is
1 y(0) 1 A 150e
k*46
2 2
75 150e
46k
applying, ln
ln 75
150
k 1
46k
* ln 1 0.015
46 2
Hence the amount of amitriptyline that is present in the blood stream of a person for
whom the half-line is 46 hours , after 24 hours is
y(24) 150e
0.015*24
150e
0.361642
104.47975
Thus the amount of amitriptyline that is present in the blood stream of a person for whom
the half-line is 31 hours , after 24 hours is 87.62 mg and that 104.47975mg for half-life
46hours
48035-8.1-56E
AID:114681 | 06/05/2018
We have an initial information wherein it’s given that half-life of Pr ozac
R is 2 3days .
We need to find out the percentage of PR that would remain after 2weeks 14days ,
considering PR gms of Pr ozacR are in the blood stream of a person for whom the half-
line is 2 days. We also need to find out the percentage that would remain after 14days,
considering PR gms of Pr ozacR are in the blood stream of a person for whom the half-
line is2 days. We need to compare these conditions.
If we let y(t) represent the ‘some quantity’ at time t, then the rate of change of ‘some
quantity’ with respect to time is y' (t) . Thus, since y' (t) is proportional to y(t) , we have
y' (t) k y(t )
For some constant of proportionality k (the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k =growth constant t=time
Considering the equation
We have
y(t) Aekt ------ (ii)
y(0) Aek*0 A PR (given) -----
(i)
The value of the growth constant, considering half-life as 2 days is
1 y(0) 1 A PRek*2 , we get the values from (i)and (ii)
2 2
Therefore 1 PR
*
2 PR
e
k*2
, half-life amount is half of the initial amount
Thus, applying natural logs on both sides, we get
ln 1 2 * k , now we know that ln 1 ln 2
2 2
Finally
Thus the amount of Pr ozacR
k
ln 2
0.69
0.34657 ------
(iii)
2 2
that is present in the blood stream of a person for whom the
half-line is 2days , after 14 days is = y(14) PRe0.34657*14 PR *0.0078
Therefore the percentage left is 100* 0.0078PR
PR 0.78%
Again, the value of the growth constant, considering half-life as 3 days is
1 y(0) 1 PR PRek*3
2 2
1
P
R
*
2
P
R
e3*k
Applying, ln
1 ln 2 k
3
k 0.23
Hence the amount of Pr ozacR
that is present in the blood stream of a person for whom
the half-line is 3days , after 14days is y(14) PRe0.23*14 PRe3.23468684 PR *0.0394
Therefore the percentage left is 100* 0.0394PR
PR 3.94%
Thus the percentage of Pr ozacR that is present in the blood stream of a person for whom
the half-line is 2days , after 2 weeks is 0.78% and that 3.94% for half-life 3 days
48035-8.1-57E
AID:114681 | 06/05/2018
2
We have an initial information wherein it’s given that half-life of ertapenem is 4 hours.
The dosage is 1gm/day. We need to find an equation for the amount left after t hours and
also plot a graph for the same
If we let y(t) represent the ‘some quantity’ at time t, then the rate of change of ‘some
quantity’ with respect to time is y' (t) . Thus, since y' (t) is proportional to y(t) , we have
y' (t) k y(t )
For some constant of proportionality k (the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k =growth constant t=time
Considering the equation
We have
y(t) Aekt ------ (ii)
y(0) Aek*0 A 1(given) ------
(i)
The value of the growth constant, considering half-life as 4 hours is
1 y(0) 1 A 1ek*4 , we get the values from (i)and (ii)
2 2
Therefore 1 *1 ek*4 , half-life amount is half of the initial amount
2 1
Thus, applying natural logs on both sides, we get
ln 1 4 * k , now we know that ln 1 ln 2
Finally
2 2
k 1 ln 1 1 ln 2 ------
(iii)
4
4
Thus the amount of ertapenem that is present in the blood stream of a person for whom
1*t*ln 2
the half-line is 4hours , t hours is = y(t) 1e 4
t ln 2
e 4
In order to plot a graph, we need to first design a table with the values
t
y(t)
0
1*0*ln 2
y(0) 1e 4 1
1
1*1*ln 2 1 ln 2
y(1) 1e 4 e 4 e0.17 0.84
2
1*2*ln 2 2 ln 2
y(2) 1e 4 e 4
e
0.34657
0.7
4
1*4*ln 2 4 ln 2
y(4) 1e 4 e 4 e0.69 0.5
8
1*8*ln 2 8 ln 2
y(8) 1e 4 e 4 e1.386 0.25
16
1*16*ln 2 16 ln 2
y(16) 1e 4 e 4 e 2.77 0.06
24
1*24*ln 2 24 ln 2
y(24) 1e 4 e 4
e
4.15888
0.016
Summarizing the table to plot our graph
t
y(t)
0
1
1
0.84
2
0.7
4
0.5
8
0.25
16
0.06
24
0.016
Hence our required equation is
t ln 2
y(t) e 4
48035-8.1-58E
AID:114681 | 06/05/2018
2
We have an initial information wherein it’s given that half-life of ertapenem is 1 hours.
The dosage is 1gm, four times a day. Let’s consider that each dosage is taken after
24/4=6hours. We need to find an equation for the amount left after t hours and also plot
a graph for the same
If we let y(t) represent the ‘some quantity’ at time t, then the rate of change of ‘some
quantity’ with respect to time is y' (t) . Thus, since y' (t) is proportional to y(t) , we have
y' (t) k y(t )
For some constant of proportionality k (the growth constant)
Now, from the above equation, we can derive,
y(t) Aekt , where A=constant, k =growth constant t=time
Considering the equation
We have
y(t) Aekt ------ (ii)
y(0) Aek*0 A 1(given) ------
(i)
The value of the growth constant, considering half-life as 1 hours is
1 y(0) 1 A 1ek*1 , we get the values from (i)and (ii)
2 2
Therefore 1 * 1 ek*1 , half-life amount is half of the initial amount
2 1
Thus, applying natural logs on both sides, we get
ln 1 1* k , now we know that ln 1 ln 2
2 2
Finally k 1 ln 1 1 ln 2 0.693 ------
(iii)
1
1
Thus the amount of ertapenem that is present in the blood stream of a person for whom
1 6
*6*ln 2
the half-line is 1hours ,after 6 hours is = y(t) 1e 1 e 1 ln 2 e4.15888
0.0156
Again, another dosage is added after 6 hours, the total quantity in the blood stream at the
6th hour is 1 0.0156 1.0156
Now,
y(0) Aek*0 A 1.01586879892(given)
1 y(0) A Aek*1
2 2
therefore, 1 ek
2
Applying, ln 1 k
2
Thus the amount of ertapenem that is present in the blood stream of a person for whom
the half-line is 1hours , after 12 hours is =
1 6
*6*ln 2
y(t) 1.0156e
1
1.0156e 1 ln 2
1.0156e
4.15888
1.0156*0.0156 0.01586879892
Again, another dosage is added after another 6 hours, the total quantity in the blood
stream at the 18th hour is 1 0.01586879892 1.01586879892
Thus the amount of ertapenem that is present in the blood stream of a person for whom
the half-line is 1hours , after 18 hours is
1
y(t) 1.01586879892e
1
6 ln 2
y(t) 1.01586879892
1
y(t) 1.01586879892e
4.15888
y(t) 1.01586879892 * 0.0156
y(t) 0.0158
Again, another dosage is added after another 6 hours, the total quantity in the blood
stream at the 24th hour is 1 0.0158 1.0158
Thus the amount of ertapenem that is present in the blood stream of a person for whom
the half-line is 1hours , after 24 hours is
1
y(t) 1.0158e
1
6 ln 2
y(t) 1.0158
1
y(t) 1.0158e
4.15888
y(t) 1.0158* 0.0156
y(t) 0.0158
Thus we can conclude that the equation is
y(t) 1.0158et*ln 2 , for any time t
*6*ln 2
*6*ln 2
In order to plot a graph, we need to first design a table with the values
t
y(t)
0
y(0) 1.0158e0*ln2 1.0158
1
y(1) 1.0158e1*ln2 1.0158eln2 0.5095
2
y(2) 1.0158e2*ln2 0.2555
4
y(4) 1.0158e4*ln2 1.0158e4ln2 1.0158e4*0.69 0.064
8
y(8) 1.0158e8*ln2 1.0158e8ln2 1.0158e8*0.69 0.004
16
y(16) 1.0158e16*ln2 1.0158e16ln2 0.00016
24
y(24) 1.0158e24*ln2 1.0158e24ln2 0.00000006
Summarizing the table to plot our graph
t
y(t)
0
1.0158
1
0.5095
2
0.2555
4
0.064
8
0.004
16
0.00016
24
0.000000006
Hence our required equation is y(t) 1.0158et*ln 2
48035-8.1-59E AID:114681 | 06/05/2018
We have an initial information wherein it’s given that a bank offers to sell a bank note
which would mature after 10 years and hold a value of $10,000. We need to find the
initial investment if we need to have to receive a 8% return
Also, we need to prove $Pert at any time, t , considering P final amount, r constant
interest rate
According to the basic formulas: P(1 r )nt where
n
P principle, r rate of interest,
t time, n number of times compounded annually
Exponential rise(+)/Exponential depreciation(-)=
v(t) value at time, t , r rate of interest
Linear depreciation v(t) mt b
v(t) Ae(rise/ depreciation)*r*t where
Exponential depreciation(-)= v(t) Aer*t
v(0) 10000 Ae
r*0
A 10, 000
,considering final value as $10,000
10 years before the asset value would have been
v(10) Ae
10*0.08
$4493.29
considering, A 10, 000
, rate r 8% 0.08
we need to prove $Pert at any time, t , considering P final amount, r constant interest
rate
This means that we need to find the value of this asset whose present worth is $P, for any
time, t
According to continuous compounding,
Current value=Principle*ert
According to the present condition, current value=$P
Thus, $P=Principle*ert
So, Principle= $P $Pert , here t =time when the initial amount was invested
ert
But if we need to find the value of t =anytime, then we need to replace principle amount
with a value which is equal to the investment value at time, t
Thus, we get $4493.29as the amount which we need to pay if we wish to receive 8%
return on our investment.
Again, in general, the present value of an item $P in t years, with constant interest r is
given by $Pert
